Question
A convex lens of refractive index $1.5$ and focal length $f = 18 \text{ cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is ________.
(refractive index of water = $4/3$)
Calculating Convex Lens Focal Length Difference in Water
The Lens Maker's Formula relates the focal length ($f$) of a lens to its refractive index ($\mu$) and the radii of curvature of its surfaces ($R_1$, $R_2$):
$ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $Let the term $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ be a constant $K$ for the given lens.
Focal Length in Air
We are given the focal length in air ($f_{air}$) and the refractive index of the lens ($\mu_{lens}$):
- $f_{air} = 18 \text{ cm}$
- $\mu_{lens} = 1.5$
Using the Lens Maker's Formula for air:
$ \frac{1}{f_{air}} = (\mu_{lens} - 1) K $We can calculate the constant $K$:
$ K = \frac{1}{f_{air}(\mu_{lens} - 1)} $ $ K = \frac{1}{18 \text{ cm} \times (1.5 - 1)} $ $ K = \frac{1}{18 \times 0.5} $ $ K = \frac{1}{9} \text{ cm}^{-1} $Focal Length in Water
When the lens is immersed in water, the refractive index changes. The refractive index of water ($\mu_{water}$) is $4/3$.
The effective refractive index of the lens material with respect to water is:
$ \mu_{lens\_in\_water} = \frac{\mu_{lens}}{\mu_{water}} = \frac{1.5}{4/3} = \frac{3/2}{4/3} = \frac{9}{8} = 1.125 $Now, we find the focal length in water ($f_{water}$) using the constant $K$:
$ \frac{1}{f_{water}} = (\mu_{lens\_in\_water} - 1) K $ $ \frac{1}{f_{water}} = (1.125 - 1) \times \frac{1}{9} \text{ cm}^{-1} $ $ \frac{1}{f_{water}} = 0.125 \times \frac{1}{9} = \frac{1/8}{9} = \frac{1}{72} \text{ cm}^{-1} $Therefore, the focal length in water is $f_{water} = 72 \text{ cm}$.
Determining the Value of Alpha
The difference in focal lengths is calculated as:
Difference = $f_{water} - f_{air} = 72 \text{ cm} - 18 \text{ cm} = 54 \text{ cm}$
The problem states this difference is $\alpha \times f$ (where $f = f_{air}$):
$ 54 \text{ cm} = \alpha \times 18 \text{ cm} $Solving for $\alpha$:
$ \alpha = \frac{54}{18} $ $ \alpha = 3 $The calculated value of $\alpha$ is $3$. This value lies within the provided correct answer range.
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